A 6.10 F capacitor that is initially uncharged is connected in series with a 470

Question

A 6.10 F capacitor that is initially uncharged is connected in series with a 4700 resistor and a 502 V emf source with negligible internal resistance.

Part A

Just after the circuit is completed, what is the voltage drop across the capacitor?

Vc = ____ A

Part B

Just after the circuit is completed, what is the voltage drop across the resistor?

Vr = _____V

Part C

Just after the circuit is completed, what is the charge on the capacitor?

Q0 = _____ C

Part D

Just after the circuit is completed, what is the current through the resistor?

IR = _____A

Part E

A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities?

Answer in the order indicated. Separate your answers with commas.

VC,VR,Q,IR = ______________ V,V,C,A

Explanation / Answer

a) since there is no charge( Q=0) on capacitor just after the circuit is completed , so voltage drop:

Vc = 0 V                                                        ( Q= CV, V= Q/ C , V = 0/C = 0V)

b) since no voltage drop across the capacitor , all the voltage drop across the resistor.

VR = 502 V

c) just after the circuit is completed there will be no charge on capacitor.

Q0 = 0C

d) just after circuit is completed current through resistor.

IR = VR / R                                                              ( V= I*R,   I = V/R)

   = 502 / 4700

   = 0.1068 A

e) (i) after long time capacitor acts like a very large resistor ( or break) , So voltage drop across capacitor

VC = 502 V

(ii) since all the voltage drop across capacitor, so voltage drop across resistor

VR = 0V

(iii) charge on capacitor, Q = CV

Q = 6.10*10-6 * 502

Q = 3.062*10-3 C

(iv) since capacitor acts like a large resistor or break , so by ohm law current through resistor

IR = V/ R                                                        

   = 0 A

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